# Question 52580

Nov 20, 2017

${\text{200 g MgCl}}_{2}$

#### Explanation:

The first thing that you need to do here is to figure out how many grams of magnesium chloride would be produced by the reaction at 100% yield.

To do that, use the molar mass of magnesium chloride

3 color(red)(cancel(color(black)("moles MgCl"_2))) * "95.211 g"/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = "285.63 g"

This means that at 100%yierld, you would expect the reaction to produce $\text{285.63 g}$ of magnesium chloride.

However, you know that the reaction has a 73% yield, which means that for every $\text{100 g}$ of magnesium chloride that the reaction could theoretically produce, you only get $\text{73 g}$.

You can thus say that your reaction will produce

285.63 color(red)(cancel(color(black)("g MgCl"_2color(white)(.)"in theory"))) * overbrace(("73 g MgCl"_2 color(white)(.)"produced")/(100color(red)(cancel(color(black)("g MgCl"_2color(white)(.)"in theory")))))^(color(blue)("= 73% yield"))

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{200 g MgCl"_2color(white)(.)"produced}}}}$

The answer is rounded to one significant figure, the number of sig figs you have for the number of moles of magnesium chloride that would be produced at 100%# yield.